Transcript How To Pump A Swing? Tareq Ahmed Mokhiemer Physics Department Contents • Introduction to the swing physics and different pumping schemes • Pumping a swing from.

How To Pump A
Swing?
Tareq Ahmed Mokhiemer
Physics Department
Contents
• Introduction to the swing physics and
different pumping schemes
• Pumping a swing from a standing position
• Qualitative understanding
• Pumping from seated position
• Qualitative understanding
• Conclusion
What is meant by “pumping a
swing” ?
Repetitive change of the rider’s
position and/or orientation
relative to the suspending rod.
How to get a swing running from a
standing position?
By standing and squatting at the lowest point
.2
1
1
2
L  m(r  )  m r  m grcos( )
2
2
• The motion of the child is a modeled by
the variation with r with time  r(t)
• This is equivalent to a parametric oscillator
.
..
. .
r   2 r    g sin 
•
Let
r  r0   sin(t )
• By scanning against pumping frequencies
amplification was found to occur at ~2 g
l
•
Constant pumping frequencie  Succession of
amplification and attenuation.
0.4
0.2
50
-0.2
-0.4
100
150
200
250
0.6
0.4
0.2
-0.2
-0.4
-0.6
320
340
360
380
Another (naïve) model for r(t)
r  l0  Cos( )
1.45
1.425
1.4
1.375
1.35
1.325
-0.6
-0.4
-0.2
0.2
0.4
0.6
0.04
0.02
50
100
150
-0.02
-0.04
No Amplification !!
Expected result !
200
A more realistic model
2.4
2.2
-1
-0.5
0.5
1.8
1.6
1
0.03
0.02
0.01
50
-0.01
-0.02
-0.03
100
150
200
• For each initial velocity  a threshold for the
steepness of r(t) at θ=0.
0.001
0.0005
50
100
150
200
-0.0005
-0.001
Unexpected result !!
How does pumping occur
physically?
Two points of view
 The conservation of angular
momentum
 Conservation of energy
conservation of angular
momentum
.
.
m rB '  B '  m rB  B
2
.
B
.
 B'
2
 2
 (1  )
rB
.
1
2
2
mrB '  B ' (n)  mgr B ' (1  cos  0 ' (n))
2
.
(1  cos 0 ' (n  1))  B ' (n  1)
 .
2
(1  cos 0 ' (n))
 B ' (n)
2
(1  cos 0 ' (n  1))
 8
 (1  )
(1  cos 0 ' (n))
rB
1
0.8
0.6
0.4
0.2
2
4
6
8
10
12
14
Conservation of energy
At the highest point
At the mid-point:
Gravitational force
W  m g.x
Centrifugal force
m v2
W 
x
l
Only gravitational force
W  mg.x
1
x
2
v (n  )  v (n)(1  2 )
2
l
2
(1  cos 0 ' (n  1))
x 2
 (1  2 )
(1  cos 0 ' (n))
l
0.6
0.5
0.4
0.3
0.2
0.1
10
20
30
40
50
Another scheme for pumping from
a standing position
The swinger pumps the swing by leaning forward and backward
while standing
Pumping a swing from a seated
position
Potential Energy =
 m1l1Cos  m2 (l2Cos(   )  l1Cos )  m3 (l1Cos  l3Cos(   ))
kinetic energy =
. 2
. 2
.
.
.
. .
.
1
1
2
2
2
2
m1l1   m2 [l1   l2 (   )  2l1l2  (   ) Cos( )] 
2
2
. 2
.
.
. .
.
1
2
2
2
m3[l1   l3 (   )  2l1l3  (   )Cos( )]
2
The equation of motion
M I1g Sin(φ(t)) + N g Sin(φ(t)+θ(t))-I1φ’’(t)
–I2 (φ’’(t)+ θ’’(t))-2 I2 N θ’’(t) Cos(θ(t))-I1 N θ’’(t) Cos(θ(t))==0
A Surprise
1
0.5
100
200
300
400
-0.5
-1
Θ(t) is either 0.5 rad
when φ is gowing or -0.5
when φ is decreasing
The oscillation
grows up
linearly!!
The growth rate is proportional to the steepness of the
frequency of the swinger’s motion
1.5
1
0.5
100
200
300
400
-0.5
-1
-1.5
Θ(t) is changes between
0.7 rad and -0.7 rad
A special case:
m2 I 2  m3 I 3  0
The Lagrangian reduces to:
.
.
1 .2 1
I1   I 2 (   ) 2  Ml1 gCos( )
2
2
And the equation of motion is
..
..
( I1  I 2 )   Mgl1Sin( )   I 2 
A driven Oscillator.
0.3
0.2
0.1
10
20
30
40
50
60
70
-0.1
-0.2
-0.3
Θ(t) changes sinusoidally
Pumping occurs at approximately the natural
frequency not double the frequency.
Pumping from a seated position…
•More efficient in starting the swing from rest position
•With the same frequency of the swinger motion, the
oscillation grows faster in the seated pumping.
Conclusion
Seated position
Standing position
• Linear growth
• Exponential growth
• Driven Oscillator
• Parametric Oscillator
• Efficient in starting the
swing from rest